Dedekind-Hasse norm implies principal ideal ring

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Statement

A commutative unital ring that admits a Dedekind-Hasse norm must be a principal ideal ring.

In particular, an integral domain that admits a Dedekind-Hasse norm must be a principal ideal domain.

Definitions used

Dedekind-Hasse norm

Further information: Dedekind-Hasse norm

A Dedekind-Hasse norm on a commutative unital ring R is a function N from the nonzero elements of R to the nonnegative integers with the property that for any elements a,b \in R with b \ne 0, it is true that either a \in (b) or there exists an element r in the ideal (a,b) such that N(r) < N(b).

Principal ideal ring

Further information: Principal ideal ring

A commutative unital ring R is termed a principal ideal ring if every ideal of R is principal.

Proof

Given: A commutative unital ring R with a Dedekind-Hasse norm N. An ideal I of R.

To prove: There exists b \in I such that I = (b).

Proof: If I = 0, we can take b = 0, and we are done.

So, suppose I \ne 0. Consider the function N on the nonzero elements of I. Since N maps to a well-ordered set, there is an element b \in I \setminus \{ 0 \} such that N(b) \le N(r) for all r \in I \setminus \{ 0 \}.

Pick any a \in I. If a = 0, then a \in (b) or there exists r \in (a,b) with N(r) < N(b).

Note that in the latter case, we have r \in (a,b) \subseteq I, so r \in I, and N(r) < N(b), contradicting the assumption that b has minimum norm among the nonzero elements of I. Hence, we have the case a \in (b). Thus, I \subseteq (b). Conversely, we clearly have (b) \subseteq I, so I = (b).